Question

Which one of the following will give a white precipitate with \[AgN{{O}_{3}}\] in an aqueous medium
A. \[\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Cl \right]{{\left( N{{O}_{2}} \right)}_{2}}\]
B. \[\left[ Pt{{\left( N{{H}_{3}} \right)}_{2}}C{{l}_{2}} \right]\]
C. \[\left[ Pt\left( en \right)C{{l}_{2}} \right]\]
D. \[\left[ Pt{{\left( N{{H}_{3}} \right)}_{4}} \right]C{{l}_{2}}\]

Answer 1

Hint: The species included in square brackets are non-ionizable and together referred to as a coordination sphere in a coordination complex. These species, known as counter ions, are present outside the coordination sphere and are ionizable. And the ionisable ions will react with $AgN{{O}_{3}}$to give the precipitate.

Complete answer:In a solution containing chloride ions, silver nitrate \[AgN{{O}_{3}}\] can precipitate by reacting with chloride ions to form \[AgCl\]. Due to its extremely poor solubility, silver chloride (AgCl), which is produced when silver nitrate combines with chloride ions, precipitates in the solution. Therefore, we will examine each of the compounds to determine which will have free chloride ions in its solution.

It can be shown in the reaction form as shown:
\[C{{l}^{-}}+AgN{{O}_{3}}\to AgCl+NO_{3}^{-}\]
If we look at option A and we write the dissociation of \[\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Cl \right]{{\left( N{{O}_{2}} \right)}_{2}}\]we will get,
\[\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Cl \right]{{\left( N{{O}_{2}} \right)}_{2}}\to {{\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Cl \right]}^{2+}}+2NO_{2}^{-}\] so here you can see that there is no free chloride ion and thus $AgN{{O}_{3}}$ will not react to form any compound or we can say it will not form \[AgCl\]and thus there will be no precipitate formation.

In option B, we are given with $[Pt{{(N{{H}_{3}})}_{2}}C{{l}_{2}}]$and we can see it’s a neutral complex so it will not give any ion in the solution and hence there will be no reaction and precipitation.

In option C, we are given with \[[Pt(en)C{{l}_{2}}]\]and that’s again a neutral ligand and will not give any ion on dissociation in solution and thus there will be no free chloride ions and hence no precipitation takes place.

If we look at option D and we write the dissociation of $[Pt{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$we will get,
$[Pt{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}\to {{[Pt{{(N{{H}_{3}})}_{4}}]}^{2+}}+2C{{l}^{-}}$and here we see the presence of free chloride ions and hence they will react with $AgN{{O}_{3}}$to form the precipitate of \[AgCl\],

The reaction can be written as
\[[Pt{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}+2AgN{{O}_{3}}\to [Pt{{(N{{H}_{3}})}_{4}}]N{{O}_{3}}+2AgCl\]here,\[AgCl\]is formed which is the white precipitate.


Hence, the correct option is D. $[Pt{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$

Note: Keep in mind that any halogen present in the solution will cause \[AgN{{O}_{3}}\] to precipitate. Therefore, precipitates would have likewise been created if bromine or iodine had been added in place of chlorine in the chemical given in option D.